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Dennis_Churaev [7]
1 year ago
5

The ideal gas law tends to become inaccurate when Group of answer choices the pressure is raised and the temperature is lowered.

the pressure is lowered and molecular interactions become significant. the temperature is raised above the temperature of STP. large gas samples are involved. the volume expands beyond the standard molar volume.
Chemistry
1 answer:
const2013 [10]1 year ago
5 0

Answer: Option (a) is the correct answer.

Explanation:

At low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas. Hence, gases behave ideally at these conditions.

Whereas at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.  

As a result, there exists force of attraction between the molecules at low temperature and high pressure and under these conditions gases are known as real gases.

Thus, we can conclude that the ideal gas law tends to become inaccurate when the pressure is raised and the temperature is lowered.

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a 75.0 liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. What is the temperature of the canister?
abruzzese [7]

Pressure of argon = 546.8 kPa

Conversion factor: 1 atm = 101.325 kPa

Pressure of argon = 546.8 kPa x 1 atm/101.325 kPa = 5.4 atm

Moles of argon = 15.82

Volume of argon = 75.0 L

According to Ideal gas law,

PV = nRT

where P is the pressure, V is the volume , n is the number of moles, R is the universal gas constant, and T is the temperature

T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol⁻¹K⁻¹)

T = 311.82 K

Hence the temperature of the canister is 311.82 K.

4 0
2 years ago
A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
Anika [276]
The heat that is required to raise the temperature of an object is calculated through the equation,
                        heat = mass x specific heat x (T2 - T1)
Specific heat is therefore calculated through the equation below,
                                specific heat = heat / (mass x (T2 - T1))
Substituting,
                                specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The value of specific heat from above equation is 0.838 J/g°C. 
5 0
2 years ago
Given that at 25.0 ∘C Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5, calculate Kb for CN− and Ka for NH4+. Enter the Kb val
neonofarm [45]

Explanation:

Using the expression :

K_a\times K_b=K_w

Where, K_w is the dissociation constant of water.

At 25\ ^0C, K_w=10^{-14}

Thus, for HCN , K_a=4.9\times 10^{-10}

<u>K_b for CN⁻ can be calculated as:</u>

K_a\times K_b=K_w

4.9\times 10^{-10}\times K_b=10^{-14}

K_b=2.0\times 10^{-5}

Thus, for NH₃ , K_b=1.8\times 10^{-5}

<u>K_a for NH_4^+ can be calculated as:</u>

K_a\times K_b=K_w

K_a\times 1.8\times 10^{-5}=10^{-14}

K_a=5.6\times 10^{-10}

5 0
2 years ago
Acetone major species present when dissolved in water
jek_recluse [69]

Answer: acetone molecule ( CH₃-CO-CH₃)


Explanation:


1) Acetone is CH₃-CO-CH₃


2) That is a molecule (build up of covalent bonds).


3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.


This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.


That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).



3 0
2 years ago
With the help of balanced chemical equation explain what happens when:(1)Zinc is put in dilute hydrochloric acid (2)Zinc is put
UkoKoshka [18]
Find the attached document.

6 0
1 year ago
Read 2 more answers
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